Answer

Verified

454.8k+ views

**Hint:**Since $|t| = 1$ we will try to separate the factor of ‘t’ from the expression $z = \dfrac{{at + b}}{{t - c}}$.

After separating the term ‘t’ we will take the modulus function of both the side and will eliminate ‘t’ using$ |t| = 1$.

**Complete step by step solution:**

Given data:

$

a,\;b,\;c,\; \text{and}\:{\text{ t}}\:{\text{are }}\:{\text{complex }}\:{\text{numbers }} \\

|t| = 1 \\

|t| \ne |c| \\

z = \dfrac{{at + b}}{{t - c}},z = x + iy. \\

$

Now solving the given equation, i.e.

$

z = \dfrac{{at + b}}{{t - c}} \\

\Rightarrow z(t - c) = at + b \\

\Rightarrow zt - at = b + cz \\

\Rightarrow t(z - a) = b + cz \\

\Rightarrow t = \dfrac{{b + cz}}{{z - a}} \\

$

Now, applying the modulus function on both sides of the equation, we obtain

$

\left| t \right| = \left| {\dfrac{{b + cz}}{{z - a}}} \right| \\

\sin ce{\text{ }}\left| t \right| = 1 \\

1 = \left| {\dfrac{{b + cz}}{{z - a}}} \right| \\

$

${\text{Now have also condition on}}$ ${\text{'c'}}$that \[\left| {\text{c}} \right| \ne {\text{1}}\]

$

1 = \left| {\dfrac{{c\left( {\dfrac{b}{c} + z} \right)}}{{z - a}}} \right| \\

we{\text{ }}know{\text{ }}that,{\text{ }}\left| {ab} \right| = \left| a \right|\left| b \right| \\

\Rightarrow 1 = \left| {\dfrac{{\dfrac{b}{c} + z}}{{z - a}}} \right|\left| c \right| \\

\Rightarrow \dfrac{1}{{\left| c \right|}} = \left| {\dfrac{{z + \dfrac{b}{c}}}{{z - a}}} \right| \\

$

Since b, c are complex numbers, then $\dfrac{b}{c}$ be also a complex number let ‘p’

$\dfrac{1}{{\left| c \right|}} = \left| {\dfrac{{z + p}}{{z - a}}} \right|$

We know that,

$\left| {\dfrac{{z - \alpha }}{{z - \beta }}} \right| = A$

When $A \ne 1$, then the locus of z is a circle.

Here also $\left| c \right| \ne 1$ i.e.,

$\left| {\dfrac{{z + p}}{{z - a}}} \right| \ne 1$

Therefore the locus of z is a circle.

**(C) The circle is the correct option.**

**Note:**It is well known that $\left| z \right| = a,$ ‘a’ being a constant is an equation of a circle.

From the equation

$z = \dfrac{{at + b}}{{t - c}}$

Applying the modulus function on both sides of the equation, we obtain

$\left| z \right| = \left| {\dfrac{{at + b}}{{t - c}}} \right|$

We can say that the right-hand side will come to be a constant as a, b, c and t are some fixed complex number.

We can write it as

$\left| z \right| = f$, where f is constant, comes out to be an equation of a circle.

Recently Updated Pages

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Which one of the following places is not covered by class 10 social science CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Who was the Governor general of India at the time of class 11 social science CBSE

How do you graph the function fx 4x class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

Difference Between Plant Cell and Animal Cell